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3 years ago

An object speeding up, an object slowing down, and an object changing direction are all examples of acceleration.

Physics

1 answer:

emmainna [20.7K]3 years ago

6 0

<h2>Answer: True</h2>

When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration. In Example B, the object is moving in the negative direction (i.e., has a negative velocity) and is slowing down.

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The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps an

Marrrta [24]

Answer: $k=55.590 KN/m$

Explanation:

Given

mass of person $\left ( m\right )$ =68 kg

car dips about 1.2 cm

We know

F=kx

Where k=combined spring constant

mg=kx

$k=\frac{mg}{x}$

$k=\frac{68\times 9.81}{1.2\time 10^{-2}}$

$k=55.590 KN/m$

7 0

4 years ago

A regulation basketball has a 15 cm diameter and may be appropriated as a thin spherical shell.

REY [17]

Solution :

From the given data,

For the spherical shell is $\frac{k^2}{R^2}= \frac{2}{3}$

where x is the radius of gyration and the acceleration of a rolling body on an inclined plane = a

Therefore,

$a =\frac{g \sin \theta}{1+ \frac{k^2}{R^2}}$

$= \frac{g \sin \theta}{1 +\frac{2}{3}}$

$= \frac{3}{5} g \sin \theta$

= 0.6 x 9.81 x sin ( 29.7)

$= 2.913 \ m/s^2$

$h = \frac{1}{2} a(\Delta t)^2$

$\Delta t = \sqrt{\frac{2h}{a}}$

$\Delta t = \sqrt{\frac{2 \times 2.9}{2.913}}$

Δt = 1.411 s

6 0

3 years ago

PLEASE HELP ME!!!

klasskru [66]

1.velocity and acceleration

3.inertia

5.speed

4 0

3 years ago

A cement truck of mass 14,000 kg moving 10m/s slams into a wall and comes to a halt in .2s. What is the force of impact on the t

skelet666 [1.2K]

And it’s x10 is 100000

6 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

4 years ago